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46

th

International Chemistry Olympiad

July 25, 2014

Hanoi, Vietnam

THEORETICAL EXAMINATION

WITH ANSWER SHEETS GRADING

Country:

Name as in passport: 
Student Code:

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GENERAL INTRODUCTION

You have additional 15 minutes to read the whole set.

This booklet is composed of 9 problems. You have 5 hours to fulfill the 
problems. Failure to stop after the STOP command may result in zero points for
the current task.

Write down answers and calculations within the designated boxes. Give your
work where required.

Use only the pen and calculator provided.

The draft papers are provided. If you need more draft paper, use the back side of
the paper. Answers on the back side and the draft papers will NOT be marked.

There are 52 pages in the booklet including the answer boxes, Cover Sheet and
Periodic Table.

The official English version is available on demand for clarification only.

Need to go to the restroom – raise your hand. You will be guided there.

After the STOP signal put your booklet in the envelope (do not seal), leave at

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Physical Constants, Units, Formulas and Equations 
Avogadro's constant NA = 6.0221 × 1023 mol–1
Universal gas constant R = 8.3145 J·K–1·mol–1
Speed of light c = 2.9979 × 108 m·s–1
Planck's constant h= 6.6261 × 10–34 J·s 
Standard pressure p° = 1 bar = 105 Pa

Atmospheric pressure 1 atm = 1.01325 × 105 Pa = 760 mmHg 
Zero of the Celsius scale 273.15 K

Mass of electron me = 9.1094 × 10–31 kg

1 nanometer (nm) = 10–9 m ; 1 angstrom (Å) = 10–10 m
1 electron volt (eV) = 1.6022 × 10–19 J = 96485 J·mol–1

Energy of a light quantum with
wavelength λ

E = hc / λ

Energy of one mole of photons Em = hcNA / λ

Gibbs energy G = H – TS

Relation between equilibrium constant

and standard Gibbs energy = exp
G
K

RT

⎛ Δ ⎞

⎜− ⎟

⎜ ⎟

⎝ ⎠

van’t Hoff equation in integral form ⎟⎟

⎠
⎞
⎜⎜

⎝
⎛

−
Δ

2
1
0
1

2 1 1

T
T
R
H
K

K

Relationship between internal energy,

heat and work ∆U = q + w

Molar heat capacity at constant volume

v
m

v

dT
dU

C ⎟

⎠
⎞
⎜
⎝
⎛
=
,

Change in internal energy from T1 to T2
assuming constant Cv,m

U(T2)=U(T1)+nCv,m(T2–T1),
Spin only formula relating number of

unpaired electrons to effective magnetic
moment

B.M.
)
2
( +

= n n

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Problem 1. Particles in a box: polyenes

In quantum mechanics, the movement of π electrons along a neutral chain of
conjugated carbon atoms may be modeled using the ‘particle in a box’ method. The
energy of the π electrons is given by the following equation:

2
2
2

8mL
h
n
En =

where n is the quantum number (n = 1, 2, 3, …), h is Planck’s constant, m is the mass
of electron, and L is the length of the box which may be approximated by L = (k +
2)×1.40 Å (k being the number of conjugated double bonds along the carbon chain in
the molecule). A photon with the appropriate wavelength λ may promote a π electron
from the highest occupied molecular orbital (HOMO) to the lowest unoccupied
molecular orbital (LUMO). An approximate semi-empirical formula based on this
model which relates the wavelength λ, to the number of double bonds k and constant B
is as follows:

λ (nm) = B

)
1

2
(

)
2

( 2

+
+
×

k
k

Equation 1

1. Using this semi-empirical formula with B = 65.01 nm calculate the value of the
wavelength λ (nm) for octatetraene (CH2 = CH – CH = CH – CH = CH – CH = CH2).

1. From the given semi-empirical formula, the wavelength λ (nm) is
calculated as follows:

)
1
2
(

)
2

(
01
.
65
)
(

2
+
+
×
=

k
k
nm

For octatetraene molecule, with k = 4; λ = 260.0 nm 3 points

Code: Question 1 2 3 4 5 Total

Examiner Mark 3 7 6 4 7 27

Theoretical

Problem 1 
5.0 % of the

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2. Derive Equation 1 (an expression for the wavelength λ (nm) corresponding to the
transfer of an electron from the HOMO to the LUMO) in terms of k and the
fundamental constants, and hence calculate theoretical value of the constant Bcalc..

2. The formula: 2 22

8mL
h
n

E= (1)

ΔE is calculated as:

ν hc

h
E

E

E= LUMO− HOMO = =

Δ (2)

In which,λ and ν are wavelength and frequency for the corresponding
photon respectively, k is the quantum number for the HOMO, which is equal
to the number of double bonds. So, we have:

(3)

]
1
2
[
8
]
)
1
[(
8 2
2
2
2
2
2
+
=
=
−
+
=
Δ k
mL
h
hc
k
k

mL
h
E
λ

Replace L = (k + 2) × 1.40 Å into (3):

2
10
2
]
10
40
.
1
)
2
[(
8
)
1
2
(
−
×
×
+
+
=
k

m
k
h
hc

λ (2 1)

]
10
40
.
1
)
2
[(

8 10 2

+
×
×
+
=
⇒ −
k
h
k
mc
λ
)

1
2
(
)
2
(
10
6261
.
6
)
10
40
.
1
(
10
9979
.
2
10
1094
.
9
8 2
34
2
10
8
31

+
+
×
×
×
×
×
×
×
×
=
⇒ − − −
k
k
λ
)
1
2
(
)
2
(
10
462
.
6
)
(
2
8

+
+
×
×
=
⇒ −
k
k
m

λ ; (4)

)
1
2
(
)
2
(
62
.
64
)
(
2
+
+
×
=
⇒

k
k
nm
λ
Bcalc. = 64.6 nm

5 points

2 points

3. We wish to synthesize a linear polyene for which the excitation of a π electron from
the HOMO to the LUMO requires an absorption wavelength of close to 600 nm. Using
your expression from part 2, determine the number of conjugated double bonds (k) in
this polyene and give its structure. [If you did not solve Part 2, use the semi-empirical 
Equation 1 with B = 65.01 nm to complete Part 3.]

3. With λ = 600 nm, we have

0
285
.
5
57
.
14
285
.
9

)
2
(

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Thus, k = 15.

So, the formula of polyene is:

CH2 = CH – (CH = CH)13 – CH = CH2 2 points

4. For the polyene molecule found in Part 3, calculate the difference in energy
between the HOMO and the LUMO, ΔE, (kJ·mol–1).

In case Part 3 was not solved, take k = 5 to solve this problem.

[

2 2

]

2
2

)
1
(

8mL k k

h
E

E

E= LUMO− HOMO = + −
Δ
⎥
⎦
⎤
⎢
⎣
⎡
+
+
×
×
×
×
×
×
×
×
×
=

Δ −34 2−31 −3 −10 223 2

)
2
(
1
2
)

10
40
.
1
(
10
1094
.
9
8
10
022
.
6
10
)
10
6261
.
6
(
k
k

E (kJ·mol–1)

⎥
⎦
⎤
⎢

⎣
⎡
+
+
×
=
Δ 2
)
2
(
1
2
1851
k
k

E (kJ/mol)

For polyene with k = 15 ; ΔE = 199 kJ·mol–1.

Taking the value of k = 5; ΔE = 415 kJ·mol–1 4 points

5. The model for a particle in a one-dimensional box can be extended to a three
dimensional rectangular box of dimensions Lx, Ly and Lz, yielding the following

expression for the allowed energy levels:

⎟

⎠

⎞

⎜

⎝

⎛

+

=

2 22

2
2
2
2
,
,

8

z
z
y
y
x
x
n
n
n

L

n

L

n

L

n

m

h

E

x y z

The three quantum numbers nx, ny, and nz must be integer values and are independent

of each other.

5.1 Give the expressions for the three different lowest energies, assuming that the box
is cubic with a length of L.

2
2
2
2
2
8
)
(
;
mL
n
n
n
h
E
L

L

Lx = y = z xyz = x + y + z

2
2
2
2
2
2
2
111
8
3
8
)
1
1
1
(
mL
h
mL
h

E = + + =

1 point

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211

121
2
2
2

2
2
2
2

112 8

6
8

)
2
1
1
(

E
E
mL

h
mL

h

E = + + = = =

221
212
2
2
2

2
2
2
2
122

8
9
8

)
2
2
1
(

E
E
mL

h
mL

h

E = + + = = =

1 point

5.2 Levels with the same energy are said to be degenerate. Draw a sketch showing all
the energy levels, including any degenerate levels, that correspond to quantum
numbers having values of 1 or 2 for a cubic box.

E111: only a single state.

E112: triple degenerate, either nx, ny or nz can equal to 2.

E122: triple degenerate, either nx, ny or nz can equal to 1.

E222: single state.

Energy diagram:

Cubic box

4 pts

E222

E122 These are no longer degenerate

E112 These are no longer degenerate

E111

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Problem 2.Dissociating Gas Cycle

Dininitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide:

N2O4(g) ⇌ 2NO2(g)

1.00 mole of N2O4 was put into an empty vessel with a fixed volume of 24.44 dm3.

The equilibrium gas pressure at 298 K was found to be 1.190 bar. When heated to
348 K, the gas pressure increased to its equilibrium value of 1.886 bar.

1a.Calculate ∆G0 of the reaction at 298K, assuming the gases are ideal.

1b. Calculate ∆H0 and ∆S0 of the reaction, assuming that they do not change
significantly with temperature.

1a. N2O4 ⇌ 2 NO2

Initial molar number 1 0
At equilibrium 1 - x 2x
ntotal,equi. = 1 - x + 2x = 1 + x (mol)

Ptotal,equi = Pini(1 + x)

(Pini - initial pressure; ntotal,equi. – total molar number of gases at equilibrium;

Ptotal,equi - total pressure of gases at equilibrium; x – number of moles N2O4

dissociated).

(mol)
174
.
1
K)

298
)(
K

J
3145
.
8
(

dm
1000

m
1
)
dm
44
.

24
(
bar
1

Pa
10
bar)
190
.
1
(

1
1

-3
3
3

, ⋅ ⋅ =

⎟⎟
⎠
⎞
⎜⎜

⎝
⎛
⎟⎟

⎠
⎞
⎜⎜

⎝
⎛
=

= −

mol
RT

PV
ntotalequi

1.174 = 1 + x

4pts

Code: Question 1a 1b 2 3 Total

Examiner Mark 12 8 3 10 33

Theoretical

Problem 2 
5.0 % of the

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x = 0.174 (mol)

∆G0 at 298 K
At equilibrium:
bar
837
.
0
bar)
190
.
1
(
174
.
0
1
174
.
0
1
1
1
4

2 + × =

−
=
×
+
−
= total
O

N x P

x
P
bar
353
.
0
bar)
190
.
1
(
174
.
0
1
174
.
0
2

1
2

2 + × =

×
=
×
+
= total
NO P
x
x
P
1489
.
0
1
837
.
0
1
353
.
0 2
0
2
0
298
4

2
2
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
P

P
P
P
K
O
N
NO

At 298 K,

)
mol
kJ
(
72
.
4
)
(
4719
)
1489
.
0
ln(
298
3145
.
8

ln 1 -1

298

0 =− =− × × = ⋅ = ⋅

ΔG RT K J mol−

1b. ∆G0 at 348 K

(mol)
593
.
1
K)
348
)(
K
J
3145
.
8
(
dm
1000
m
1
)
dm
44

.
24
(
bar
1
Pa
10
bar)
886
.
1
(
1
1

-3
3
3
5
, ⋅ ⋅ =
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜

⎝
⎛
=
= −
mol
RT
PV
ntotalequi

1.593 = 1 + x
x = 0.593 (mol)
At equilibrium:
bar
482
.
0
bar)
886
.
1
(
593
.
0
1
593
.
0
1
1

1
4

2 + × =

−
=
×
+
−
= total
O
N P
x
x
P
bar
404
.
1
bar)
886
.
1
(
593
.
0
1
593

.
0
2
1
2

2 + × =

×
=
×
+
= total
NO P
x
x
P
0897
.
4
1
482
.
0
1
404
.
1 2
0
2

0
348
4
2
2
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝

⎛
=
⇒
P
P
P
P
K
O
N
NO

At 348 K,

4pts

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∆S0

∆G0348 = - 4.07 kJ = ∆H – 348∆S (1)
∆G0298 = 4.72 kJ = ∆H – 298∆S (2)

(2) - (1) →∆S = 0.176 kJ·mol–1·K–1

∆H0

∆H0 = 4.720 + 298 × 0.176 = 57.2 (kJ·mol–1)

4pts

If you cannot calculate ∆H0, use ∆H0= 30.0 kJ·mol–1 for further calculations.

The tendency of N2O4 to dissociate reversibly into NO2 enables its potential use in

advanced power generation systems. A simplified scheme for one such system is
shown below in Figure (a). Initially, "cool" N2O4 is compressed (1→2) in a

compressor (X), and heated (2→3). Some N2O4 dissociates into NO2. The hot mixture

is expanded (3→4) through a turbine (Y), resulting in a decrease in both temperature
and pressure. The mixture is then cooled further (4→1) in a heat sink (Z), to promote
the reformation of N2O4. This recombination reduces the pressure, thus facilitates the

compression of N2O4 to start a new cycle. All these processes are assumed to take

place reversibly.

X

Y

1

q in

work out

2 3 4

4

q out

Z

(a)

To understand the benefits of using reversible dissociating gases such as N2O4, we will

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2. Give the equation to calculate the work done by the system w(air) during the

reversible adiabatic expansion for 1 mol of air during stage 3 → 4. Assume that
Cv,m(air) (the isochoric molar heat capacity of air) is constant, and the temperature

changes from T3 to T4.

∆U = q + w; work done by turbine w(air)=-w 1 pt
q = 0, thus w(air) = ∆U = Cv,m(air)[T3-T4] 2 pts

3.Estimate the ratio w(N2O4)/w(air), in which w(N2O4) is the work done by the gas during

the reversible adiabatic expansion process 3 → 4 with the cycle working with 1 mol of
N2O4, T3 and T4 are the same as in Part 2. Take the conditions at stage 3 to be T3 = 440

K and P3 = 12.156 bar and assume that:

(i) the gas is at its equilibrium composition at stage 3;
 (ii) Cv,m for the gas is the same as for air;

(iii) the adiabatic expansion in the turbine takes place in a way that the
composition of the gas mixture (N2O4 + NO2) is unchanged until the expansion is

completed.
⎟
⎠
⎞
⎜
⎝
⎛ −
=
⎟
⎠
⎞
⎜
⎝
⎛ −
Δ
=
440
1
348
1
3145
.
8
57200
440
1
348

1
ln 0
348
440
R
H
K
K
542
.
5
440
1
348
1
314
.
8
57200
0897
.
4
ln
440
1
348
1
3145
.
8

57200
ln

ln 440 348 ⎟=

⎠
⎞
⎜
⎝
⎛ −
×
+
=
⎟
⎠
⎞
⎜
⎝
⎛ −
×
+
= K
K

→K440 = 255.2

N2O4 ⇌ 2 NO2 (1)

Initial molar number 1 0
At equilibrium 1 - x 2x

ntotal = 1 - x + 2x = 1 + x (mol); Ptotal = 12.156 bar

At equilibrium: 12.156(bar)
1

1
4

2 + ×

−
=

x
x

PNO ; 12.156(bar)

1
2
2 = +x×

x
PNO

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2
.
255
1
156

.
12
1
1
1
156
.
12
1
2 2
0
2
0
440
4
2
2
=
⎟⎟
⎟
⎟
⎠
⎞
⎜⎜
⎜
⎜
⎝
⎛ ×
+
−

⎟⎟
⎟
⎟
⎠
⎞
⎜⎜
⎜
⎜
⎝
⎛ ×
+
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
⇒
x
x
x
x

P
P
P
P
K
O
N
NO

(P0 = 1 bar) → = ⇒

−
⇒
=
⎟
⎠
⎞
⎜
⎝
⎛
+
−
⎟
⎠
⎞
⎜
⎝
⎛
+

99
.
20
1
4
99
.
20
1
1
1
2
2
2
2
x
x
x
x
x
x

4x2 = 20.99 – 20.99 x2

→ 24.99 x2 = 20.99 → x = 0.92; ntotal = 1 + x = 1.92
→ wN2O4 = 1.92 × Cv,air × (T3 – T4); → 2 4 =1.92

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Problem 3. High-valent Silver Compounds

Silver chemistry is dominated by Ag (I) compounds. Compounds of silver in higher

oxidation state (from +2 to +5) are not very abundant due to their instability with
respect to reduction. High-valent silver compounds are very reactive and can be
synthesized from Ag(I) compounds in electro-chemical oxidations or in chemical
oxidations using powerful oxidizing agents.

1. In some peroxydisulfate (S2O82-) oxidations catalyzed by Ag+, black solid (A) with the

composition AgO can be isolated.

1a. Choose the appropriate magnetic behaviour of A if it exists as AgIIO.

Diamagnetic Paramagnetic

x 1 point

Single crystal X - ray studies reveal that the lattice of A contains two nonequivalent Ag
atom sites (in equal proportions) of which one denoted as Ag1 and the other denoted as
Ag2. Ag1 shows a linear O atom coordination (O-Ag-O) and Ag2 shows a square-planar
O atom coordination. All O atoms are in equivalent environments in the structure. Thus,

A should be assigned as AgIAgIIIO2 rather than AgIIO.
1b. Assign the oxidation number of Ag1 and Ag2.
Oxidation number of Ag1 : ……….+1

Oxidation number of Ag2 : ……… +3 2 points

Code: Question 1 2 3 4 Total

Examiner Marks 8 14 2 12 36

Theoretical

Problem 3

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1c.What is the coordination number of O atoms in the lattice of A?

The coordination number of O atoms =……… 3 1 point
1d.How many AgI and AgIII bond to one O atom in the lattice of A?
Number of AgI = ……… 1

Number of AgIII = ……. 2 2 points

1e.Predict the magnetic behaviour of A. Check the appropriate box below.

Diamagnetic Paramagnetic

x 1 point

The AgI is d10 hence diamagnetic; the AgIII is square planar d8 also diamagnetic

1f. The compound A can also be formed on warming a solution of Ag+ with
peroxydisulfate. Write down the equation for the formation of A.

S2O82-(aq) + 2Ag+(aq) + 2H2O (l) 2SO42-(aq) + AgIAgIIIO2 (s) + 4H+(aq)

1 point

2. Among the silver oxides which have been crystallographically characterized, the
most surprising is probably that compound A is not a AgIIO. Thermochemical cycles
are useful to understand this fact. Some standard enthalpy changes (at 298 K) are

listed:

Atom

Standard enthalpy
of formation

(kJ·mol–1)

1st ionization
(kJ·mol–1)

2ndionization
(kJ·mol–1)

3rd ionization
(kJ·mol–1)

1st electron
affinity
(kJ·mol–1)

2nd electron
affinity
(kJ·mol–1)

Cu(g) 337.4 751.7 1964.1 3560.2
Ag(g) 284.9 737.2 2080.2 3367.2

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Compounds ΔHof (kJ·mol–1)

AgIAgIIIO2 (s) –24.3

CuIIO (s) –157.3

The relationship between the lattice dissociation energy (Ulat) and the lattice 
dissociation enthalpy (ΔHlat) for monoatomic ion lattices is: ΔHlat =Ulat +nRT, where n 
is the number of ions in the formula unit.

2a. Calculate Ulat at 298 K of AgIAgIIIO2 and CuIIO. Assume that they are ionic

compounds.

Ulat of AgIAgIIIO2

Calculations:

ΔHlat (AgIAgIIIO2) = 2 ΔHof (O2-) + ΔHof (Ag+) + ΔHof (Ag3+) –ΔHof (AgIAgIIIO2)

= (2×249 – 2 × 141 + 2 × 844) + (284.9 + 737.2) + (284.9 + 737.2
+ 2080.2 + 3367.2 ) – (–24.3)

= +9419.9 (kJ·mol–1)

U lat (AgIAgIIIO2) = ΔHlat (AgIAgIIIO2) – 4RT

= + 9419.9 – 10.0 = + 9409.9 (kJ·mol–1) 3 points
(no penalty if negative sign)
Ulat of CuIIO

Calculations for: Ulat of CuIIO

ΔHlat (CuIIO) = ΔHof (O2–) + ΔHof (Cu2+) – ΔHof (CuIIO)

= (249 – 141 + 844) + (337.4 + 751.7 + 1964.1) – (–157.3)
= 4162.5 (kJ·mol–1)

U lat (CuIIO) = ΔHlat (CuIIO) – 2RT = 4162.5 – 5.0 = 4157.5 (kJ·mol–1) 
3 points
(no penalty if negative sign)

If you can not calculate the Ulat of AgIAgIIIO2 and CuIIO, use following values for

further calculations: Ulat of AgIAgIIIO2 = 8310.0 kJ·mol–1; Ulat of CuIIO = 3600.0

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The lattice dissociation energies for a range of compounds may be estimated using this
simple formula:

3
1

C ⎟⎟

⎠
⎞
⎜⎜
⎝
⎛

×
=

m
lat

V
U

Where: Vm (nm3) is the volume of the formula unitand C (kJ·nm·mol–1) is an empirical

constant which has a particular value for each type of lattice with ions of specified
charges.

The formula unit volumes of some oxides are calculated from crystallographic data as
the ratio between the unit cell volume and the number of formula units in the unit cell
and listed as below:

Oxides Vm (nm3)

CuIIO 0.02030

AgIII2O3 0.06182

AgIIAgIII2O4 0.08985

2b.CalculateUlat for the hypothetical compound AgIIO. Assume that AgIIO and CuIIO

have the same type of lattice, and that Vm (AgIIO) = Vm (AgIIAgIII2O4) – Vm (AgIII2O3).

Calculations:

Vm (AgIIO) = Vm (AgIIAgIII2O4) - Vm (AgIII2O3) = 0.08985 – 0.06182 = 0.02803 nm3

From the relationship Ulat = C×(Vm)–1/3 we have

Ulat (AgIIO) = 3

02803
.
0

02030
.
0
5
.

4157 × = 3733.6 (kJ·mol-1) 3 points

Answer: 3733.6 (kJ.mol-1) [or 3232.9 kJ·mol–1 if using Ulat CuIIO = 3600 kJ·mol-1]

2c. By constructing an appropriate thermodynamic cycle or otherwise, estimate the
enthalpy change for the solid-state transformation from AgIIO to 1 mole of AgIAgIIIO2.

(Use Ulat AgIIO = 3180.0 kJ·mol-1 and Ulat AgIAgIIIO2 = 8310.0 kJ·mol-1 if you cannot

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2AgIIO(s) AgIAgIIIO2 (s)

2Ag2+(g) + 2O2-(g) Ag+(g) + Ag3+(g) + 2O2-(g)

Hrxn

2Ulat(AgO) + 4RT - Ulat(AgIAgIIIO) - 4RT

IE3(Ag) -IE2(Ag)

Calculations:

ΔHrxn = 2Ulat (AgIIO) + 4RT + IE3 – IE2 – Ulat (AgIAgIIIO2) – 4RT

= 2 × 3733.6 + 3367.2 – 2080.2 – 9409.9

= – 655.7 (kJ/mol) or - 663.0 kJ/mol using given Ulat values 4 pts

2d. Indicate which compound is thermodynamically more stable by checking the

appropriate box below.

AgIIO AgIAgIIIO2

x 1 point

3. When AgIAgIIIO2 is dissolved in aqueous HClO4 solution, a paramagnetic

compound (B) is first formed then slowly decomposes to form a diamagnetic
compound (C). Given that B and C are the only compounds containing silver formed
in these reactions, write down the equations for the formation of B and C.

For B:

AgIAgIIIO2(s) + 4 HClO4(aq) 2Ag(ClO4)2(aq) + 2 H2O (l) 1 point

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4. Oxidation of Ag+ with powerful oxidizing agents in the presence of appropriate
ligands can result in the formation of high-valent silver complexes. A complex Z is
synthesized and analyzed by the following procedures:

An aqueous solution containing 0.500 g of AgNO3 and 2 mL of pyridine (d =

0.982 g/mL) is added to a stirred, ice-cold aqueous solution of 5.000 g of K2S2O8. The

reaction mixture becomes yellow, then an orange solid (Z) is formed which has a mass
of 1.719 g when dried.

Elemental analysis of Z shows the mass percentages of C, H, N elements are
38.96%, 3.28%, 9.09%, respectively.

A 0.6164 g Z is added to aqueous NH3. The suspension is boiled to form a clear

solution during which stage the complex is destroyed completely. The solution is
acidified with excess aqueous HCl and the resulting suspension is filtered, washed and
dried (in darkness) to obtain 0.1433 g of white solid (D). The filtrate is collected and
treated with excess BaCl2 solution to obtain 0.4668 g (when dry) of white precipitate

(E).

4a. Determine the empirical formula of Z and calculate the percentage yield in the
preparation.

Calculations:

- Mole Agin 0.6164 g of Z = mole of AgCl = 0.001 mole
- Mole SO42- from 0.6160 g of Z = mole BaSO4 = 0.002 mol

- Mass percentage of Ag = 0.001×107.87/0.6164 = 17.50 %
- Mass percentage of SO42- = 0.002×96.06/0.6164 = 31.17 %

- From EA:

Ratio Ag2+ : SO42- : C : H : N =

01
.
14

09
.
9
:
01
.
1

28
.
3
:
01
.
12

96
.
38
:
12
.
192

17
.
31
:
87
.
107

50
.
17

= 1 : 2 : 20 : 20: 4
The empirical formula of Z is: C20H20AgN4O8S2 2 points

Yield = 100%

4
.
616
169.87

0.500

1.719 ×

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4b. Ag (IV) and Ag (V) compounds are extremely unstable and found only in few
fluorides. Thus, the formation of their complexes with organic ligands in water can be
discounted. To confirm the oxidation number of silver in Z, the effective magnetic
moment (µeff ) of Z was determined and found to be 1.78 BM. Use the spin only

formula to determine the number of unpaired electrons in Z and the molecular
formula of Z. (Z contains a mononuclear complex with only one species of Ag and
only one type of ligand in the ligand sphere.)

- n(n+2) =1.78 (n is number of unpaired electron of Ag)
- n = 1, corresponds to AgII (d9)

- Most rational molecular formula of Z is [AgII(Py)4](S2O8) 3 point
4c. Write down all chemical equations for the preparation of Z, and its analysis.
Formation of Z:

2Ag+(aq) + 8Py (l) + 3S2O82–(aq) 2[AgII(Py)4](S2O8) (s) + 2SO42–(aq) 2 pts
Destruction of Z with NH3:

[AgII(Py)4](S2O8) (s) + 6NH3(l) [Ag(NH3)2]+(aq) + ½ N2(g) + 2SO42-(aq)+3NH4+
(aq) + 4Py (l) 2 pts

(All reasonable N –containing products and O2 are acceptable)
Formation of D:

[Ag(NH3)2]+(aq) + 2H+(aq) + Cl– (aq) AgCl (s) + 2NH4+(aq) 1 pt

Formation of E:

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Problem 4. Zeise’s Salt

1. Zeise's salt, K[PtCl3C2H4], was one of the first organometallic compounds to be

reported. W. C. Zeise, a professor at the University of Copenhagen, prepared this
compound in 1827 by reacting PtCl4 with boiling ethanol and then adding potassium

chloride (Method 1). This compound may also be prepared by refluxing a mixture of
K2[PtCl6] and ethanol (Method 2). The commercially available Zeise's salt is

commonly prepared from K2[PtCl4] and ethylene (Method 3).

1a.Write balanced equations for each of the above mentioned preparations of Zeise's
salt, given that in methods 1 and 2 the formation of 1 mole of Zeise’s salt consumes 2
moles of ethanol.

PtCl4 + 2 C2H5OH → H[PtCl3C2H4] + CH3CH=O + HCl + H2O

H[PtCl3C2H4] + KCl → K[PtCl3C2H4] + HCl

K2[PtCl6] + 2 C2H5OH → K[PtCl3C2H4] + CH3CH=O + KCl + 2 HCl + H2O

K2[PtCl4] + C2H4 → K[PtCl3C2H4] + KCl

1pt for each (2 pts if the first two reactions combined), total of 4 pts

1b. Mass spectrometry of the anion [PtCl3C2H4]– shows one set of peaks with mass

numbers 325-337 au and various intensities.

Calculate the mass number of the anion which consists of the largest natural
abundance isotopes (using given below data).

Code: Question 1a 1b 2a 3a 3b 3c Total

Examiner Mark 4 1 10 2 6 4 27

Theoretical

Problem 4

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Isotope 192Pt

78 Pt
194

78 Pt
195

78 Pt
196

78 Pt
198

78 C

6 C

13
6

Natural abundance,

% 0.8 32.9 33.8 25.3 7.2 75.8 24.2 98.9 1.1 99.99
Calculations:

195 + 3×35 + 2×12 + 4×1 = 328 1 pt

2. Some early structures proposed for Zeise’s salt anion were:

In structure Z1, Z2, and Z5 both carbons are in the same plane as dashed square. [You
should assume that these structures do not undergo any fluxional process by
interchanging two or more sites.]

2a. NMR spectroscopy allowed the structure for Zeise’s salt to be determined as
structure Z4. For each structure Z1-Z5, indicate in the table below how many
hydrogen atoms are in different environments, and how many different environments
of hydrogen atoms there are, and how many different environments of carbon atoms
there are?

Structure Number of different

environments of hydrogen

Number of different

environments of carbon

Z1 2

1pt

1 pt

Z2 2

1pt

1 pt

Z3 2

1pt

1 pt

Z4 1

1pt

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3. For substitution reactions of square platinum(II) complexes, ligands may be
arranged in order of their tendency to facilitate substitution in the position trans to
themselves (thetrans effect). The ordering of ligands is:

CO , CN- , C2H4 > PR3 , H- > CH3- , C6H5- , I- , SCN- > Br- > Cl- > Py > NH3 > OH- , H2O

In above series a left ligand has stronger trans effect than a right ligand.

Some reactions of Zeise’s salt and the complex [Pt2Cl4(C2H4)2] are given below.

3a.Draw the structure of A, given that the molecule of this complex has a centre of
symmetry, no Pt-Pt bond, and no bridging alkene.

Structure of A

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3b.Draw the structures of B, C, D, E, F and G.

B

1 pt

C

Cl NH2C6H5

1 pt

D

1 pt

E

1 pt

F

1 pt

G

1 pt

3c.Suggest the driving force(s) for the formation of D and F by choosing one or more
of the following statements (for example, i and ii):

i) Formation of gas
ii) Formation of liquid
iii) Trans effect

iv) Chelate effect

Structure D F

Driving force(s) i iii and iv

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Problem 5. Acid-base Equilibria in Water

A solution (X) contains two weak monoprotic acids (those having one acidic
proton); HA with the acid dissociation constant of KHA = 1.74 × 10–7, and HB with the

acid dissociation constant of KHB= 1.34 × 10–7. The solution X has a pH of 3.75.

1. Titration of 100 mL solution X requires 100 mL of 0.220 M NaOH solution for
completion.

Calculate the initial (total) concentration (mol·L–1) of each acid in the solution X.

Use reasonable approximations where appropriate. [KW = 1.00 × 10–14at 298 K.]

Solution: In solution X, H+ was produced from the reactions :

HA H+ + A– and HB H+ + B– and H2O H+ + OH–
The positive and negative charges in an aqueous solution must balance. Thus the charge
balance expression is:

[OH–] + [A–] + [B–] = [H+] (Eq.1)

In the acidic solution (pH = 3.75), [OH–] can be neglected, so:

[A–] + [B–] = [H+] (Eq. 2)

From equilibrium expression: KHA

HA
A

H+ × − =

]
[

]
[
]
[

and [HA] = [HA]i – [A–] (where [HA]i is the initial concentration)

So: [H+]×[A−]= KHA×[HA]=KHA

(

[HA]i −[A−])

)

Thus, the equilibrium concentration of [A–] can be presented as:

[ ]

]
[

+
−

+
×
=

H
K

HA
K

A

HA

i
HA

Similarly, the equilibrium concentration of [B–] can be presented as:

Code: Question 1 2 3 4 Total

Examiner Mark 6 4 4 6 20

Theoretical

Problem 5

6.5 % of the

</div>
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[ ]

]
[

+
−

+
×
=

H
K

HB
K

B

HB

i
HB

Substitute equilibrium concentrations of [A–] and [B–] into Eq.2:

[ ]

+
+

+ + =

×
+
+

H
H

K

HB
K

H
K

HA
K

HB

i

HB

HA

i
HA

]
[

]
[
]

[
]
[

2 pts

Since KHA, KHB are much smaller than [H+], thus:

[ ]

+
+

+ =

×
+
×

H
H

HB
K

H
HA

KHA i HB i

]
[

]
[
]

[

]
[

or 1.74 × 10–7 × [HA]i + 1.34 × 10–7 × [HB]i = [H+]2 = (10–3.75 )2

1.74 × [HA]i + 1.34 × [HB]i= 0.316 (Eq. 3)

Neutralization reactions show:

HA + NaOH NaA + H2O

HB + NaOH NaB + H2O

nHA + nHB = nNaOH

or ([HA]i + [HB]i) × 0.1 L = 0.220 M × 0.1 L 2 pts

[HA]i + [HB]i= 0.220 M (Eq. 4)

Solving Eq.3 and Eq.4 gives: [HA]i = 0.053 M and [HB]i = 0.167 M

Concentration of HA = 0.053 M

Concentration of HB = 0.167 M 2 pts

2. Calculate the pH of the solution Y which initially contains 6.00×10-2 M of NaA and
4.00×10-2 M of NaB.

Solution:

Solution Y contains NaA 0.06 M and NaB 0.04 M. The solution is basic, OH– was
produced from the reactions:

NaA + H2O HA + OH– Kb,A = Kw/KHA = 5.75 ×10-8

NaB + H2O HB + OH– Kb,B = Kw/KHB = 7.46 ×10-8

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[H+] + [HA] + [HB] = [OH–] (Eq. 5)
In the basic solution, [H+] can be neglected, so:

[HA] + [HB] = [OH–] (Eq. 6)

From equilibrium expression: KbA

A
HA
OH
,
]
[
]
[
]
[ × =
−
−

and [A–] = 0.06 – [HA] 1 pt
Thus, the equilibrium concentration of HA can be presented as:

[ ]

]
[
06
.
0
.
,
−
+

×
=
OH
K
K
HA
A
b
A
b

Similarly, the equilibrium concentration of HB can be presented as:

[ ]

]
[
04
.
0
.
,
−
+
×
=
OH
K
K
HB
B
b

B
b

Substitute equilibrium concentrations of HA and HB into Eq. 6:

]
[
06
.
0
.
,
−
+
×
OH
K
K
A
b
A

b +

]
[
04
.
0
.

,
−
+
×
OH
K
K
B
b
B

b = [OH–] 2 points

Assume that Kb,Aand Kb,B are much smaller than [OH–] (*), thus:

[OH–] 2 = 5.75 × 10 –8 × 0.06 + 7.46 × 10 –8 × 0.04
[OH–] = 8.02 × 10 –5 (the assumption (*) is justified)

So pOH = 4.10 and pH = 9.90 1 point

3. Adding large amounts of distilled water to solution X gives a very (infinitely) dilute
solution where the total concentrations of the acids are close to zero. Calculate the
percentage of dissociation of each acid in this dilute solution.

Solution: HA in the dilute solution:
[A–] = α × [HA]i

[HA] = (1 - α ) × [HA]i

[H+] = 10–7

Substitute these equilibrium concentrations into KHA expression:
HA
i
i K
HA
HA =
×
−
×
×
−
]
[
)
1
(
]
[
10 7
α
α

or 7 1.74 10 7

)
1
(

10− = × −

−

×
α

α 2 pts

Solving the equation gives: α = 0.635

Similarly, for HB: 7 1.34 10 7

)
1
(

10− = × −
−

×
α

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Solving the equation gives: α = 0.573

- The percentage of dissociation of HA = 65.5 %

- The percentage of dissociation of HB = 57.3 % 2 points

4. A buffer solution is added to solution Y to maintain a pH of 10.0. Assume no
change in volume of the resulting solution Z.

Calculate the solubility (in mol·L–1) of a subtancce M(OH)2 in Z, given that the anions

A– and B– can form complexes with M2+:

M(OH)2 M2+ + 2OH– Ksp = 3.10 ×10-12

M2+ + A– [MA]+ K1= 2.1 × 103

[MA]+ + A– [MA2] K2 = 5.0 × 102

M2+ + B– [MB]+ K’1 = 6.2 × 103

[MB]+ + B– [MB2] K’2 = 3.3 × 102
Solution:

M(OH)2 M2+ + 2OH– Ksp = 3.10 ×10-12

H2O H+ + OH– Kw = 1.00 × 10-14

M2+ + A– [MA]+ K1 = 2.10 × 103

[MA]+ + A– [MA2] K2 = 5.00 × 102

M2+ + B– [MB]+ K’1 = 6.20 × 103

[MB]+ + B– [MB2] K’2 = 3.30 × 102

Solubility of M(OH)2 = s = [M2+] + [MA+] + [MA2] + [MB+] + [MB2]

pH of Z = 10.0

4
2

4
12
2

2 3.10 10

)
10
(

10
10
.
3
]
[
]

[ −

−
−
−

+ = = × = ×

OH
K

M sp M Eq.1

At pH = 10.0

06
.
0
)
10
(

06
.
0
]

[ 10 =

+
×

= −

−

HA
HA

total K

K
A

[MA+] = K1[M2+][A-–] = 2.1 × 103 × 3.10 × 10–4 ×[A–] = 0.651 ×[A–] Eq. 3

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Solve this equation: [A-] = 8.42× 10 –3 M

Substitute this value into Eq. 3 and Eq. 4:
[MA+] = 0.651 × [A–] = 5.48 × 10 –3 M

[MA2] = 325.5 × [A–]2 = 2.31 × 10 –2 M

Similarly,

[B–]total = 0.04 M

]
[
92
.
1
]
[
10
10
.
3
10

2
.
6
]
][
[
]

[ ' 2 3 4

1 + − − − −

+ =K M B = × × × × B = × B

MB Eq. 6

2
2

2
'
2
'
1

2] [ ][ ] 634.3 [ ]

[MB =K K M + B− = × B− Eq.7

[B–]total = [B-] + [MB+] + 2 × [MB2] = 0.04 M Eq. 8 2pts

Substitute Eq. 6 and Eq. 7 into Eq. 8:
[B–] + 1.92 × [B–] + 2 × 634.3 × [B–]2 = 0.04
Solve this equation: [B–] = 4.58 × 10–3 M 
Substitute this value into Eq. 6 and Eq. 7:
[MB+] = 1.92 ×[B–] = 8.79 × 10 –3 M

[MB2] = 634.3 ×[B–]2 = 1.33 × 10–2 M

Thus, solubility of M(OH)2 in Z is s’

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Problem 6. Chemical Kinetics

The transition-metal-catalyzed amination of aryl halides has become one of the most
powerful methods to synthesize arylamines. The overall reaction for the
nickel-catalyzed amination of aryl chloride in basic conditions is:

in which NiLL’ is the nickel complex catalyst. The reaction goes through several steps
in which the catalyst, reactants, and solvent may be involved in elementary steps.

6a. To determine the reaction order with respect to each reactant, the dependence of
the initial rate of the reaction on the concentrations of each reagent was carried out
with all other reagents present in large excess. Some kinetic data at 298 K are shown
in the tables below. (Use the grids if you like)

[ArCl]
(M)

Initial rate
(M s–1)

0.1 1.88 × 10-5

0.2 4.13×10-5

0.4 9.42 × 10-5

0.6 1.50 × 10-4 0

2
4
6
8
10
12
14
16

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

Code: Question 6a 6b 6c 6d 6e Total

Examiner Marks 6 8 4 12 2 32

Theoretical

Problem 6 
7.0 % of the

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[NiLL’]

(M)

Initial rate
(M s–1)

6 × 10–3 4.12 × 10–5

9 × 10–3 6.01 × 10–5

1.2 × 10–2 7.80 × 10–5

1.5 × 10–2 1.10 × 10–4

0
2
4
6
8
10
12

0 0.3 0.6 0.9 1.2 1.5 1.8

[L’]
(M)

Initial rate
(M s–1)

0.06 5.8 × 10–5

0.09 4.3 × 10–5

0.12 3.4 × 10–5

0.15 2.8 × 10–5

0
1
2
3
4
5
6

0 0.03 0.06 0.09 0.12 0.15 0.18

Determine the order with respect to the reagents assuming they are integers.

- Order with respect to [ArCl] = = 1
- Order with respect to [NiLL’] = = 1

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6b. To study the mechanism for this reaction, 1H, 31P, 19F, and 13C NMR spectroscopy
have been used to identify the major transition metal complexes in solution, and the
initial rates were measured using reaction calorimetry. An intermediate, NiL(Ar)Cl,
may be isolated at room temperature. The first two steps of the overall reaction involve
the dissociation of a ligand from NiLL’ (step 1) at 50 oC, followed by the oxidation
addition (step 2) of aryl chloride to the NiL at room temperature (rt):

Using the steady state approximation, derive an expression for the rate equation for

the formation of [NiL(Ar)Cl].

The rate law expression for the formation of NiLAr(Cl)

rate =

]
)[
/
(
]
'
[

]
[
]
'
)[
/
(
]
[
]
'
[

]
[
]

'
[

1
2
1
2
1
2

1
2
1

ArCl
k

k
L

ArCl
NiLL

k
k
k
ArCl
k

L

k

ArCl
NiLL

k
k

−
−

− +

+ 8 pts

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The next steps in the overall reaction involve the amine (RNH2) and tBuONa.

To determine the order with respect to RNH2 and tBuONa, the dependence of the

initial rates of the reaction on the concentrations of these two reagents was carried with
the other reagents present in large excess. Some results are shown in the tables below.

[NaOtBu], 
(M)

Initial rate
(M·s–1)

0.2 4.16 × 10–5

0.6 4.12 × 10–5

0.9 4.24 × 10–5

1.2 4.20 × 10–5 0

0.5
1
1.5
2
2.5
3
3.5
4
4.5

0 0.2 0.4 0.6 0.8 1 1.2 1.4

[RNH2]

(M)

Initial rate
(M s–1)

0.3 4.12 × 10–5

0.6 4.26 × 10–5

0.9 4.21 × 10–5

1.2 4.23 × 10–5 0

0.5
1
1.5
2
2.5
3
3.5
4
4.5

0 0.3 0.6 0.9 1.2 1.5

</div>
(33)<div class='page_container' data-page=33>

6c. Determine the order with each of these reagents, assuming each is an integer.

(Use the grids if you like)

- Order with respect to [NaOtBu] = 0 2 pts
- Order with respect to [RNH2] = 0 2 pts

During a catalytic cycle, a number of different structures may be involved
which include the catalyst. One step in the cycle will be rate-determining.

A proposed cycle for the nickel-catalyzed coupling of aryl halides with amines is as
follows:

6d. Use the steady-state approximation and material balance equation to derive the
rate law for d[ArNHR]/dt for the above mechanism in terms of the initial
concentration of the catalyst [NiLL’]0 and concentrations of [ArCl], [NH2R],

[NaOtBu], and [L’].

Using the mechanism depicted by Reaction (1) through (4), the rate equation:

]
'
][
[
]
'
[
]
'
[
1

1 NiLL k NiL L

k
dt
NiLL
d
−
+
−
=

]
)
(
[
]
][
[
]
'
][
[
]
'
[
]
[
4
2
1

1 NiLL k NiL L k NiL ArCl k NiL Ar NHR

k
dt
NiL
d
+
−
−
= −

Apply the steady-state approximation to the concentrations for the intermediates:

0
]
[ =
dt
NiL
d

k1[NiLL’] = k-1[NiL][L’] + k2[NiL][ArCl] – k4[NiL(Ar)HNR] (Equation 1) 1pt

0
]
)
(
][
][
[
]
][
[
]
)
(
[
2
3

2 − =

</div>
(34)<div class='page_container' data-page=34>

0
]
)
(
[
]
][
][
)
(
[
]
)
(
[
4
2

3 − =

=k NiL Ar Cl NH R NaOBu k NiL Ar NHR
dt
NHR
Ar
NiL
d
]
][
][

)
(
[
]
)
(
[ 2
4

3 NiL Ar Cl NH R NaOBu

k
k
NHR
Ar

NiL = ( Equation 3)

Substitute Equation 2 into Equation 3:

]
][
[
]
][
[
]
][
[
]

][
[
]
)
(
[
4
2
2
3
2
2
4

3 ArCl NiL

k
k
NaOBu
R
NH
NiL
ArCl
k
k
NaOBu
R
NH
k
k

NHR
Ar

NiL = × = (Eq. 4) 1pt

Substitute Equation 4 into Equation 1:

]
)
(
[
]
][
[
]
'
][
[
]
'

[ 1 2 4

1 NiLL k NiL L k NiL ArCl k NiL Ar NHR

k = − + −

]
'
][

[
]
][
[
]
][
[
]
'
][
[
]
'
[ 1
4
2
4
2
1

1 ArCl NiL k NiL L

k
k
k
ArCl
NiL
k
L
NiL

k
NiLL

k = − + − × = − (Eq.5)

The material balance equation with respect to the catalyst is

[NiLL’]0 = [NiLL’] + [NiL] + [NiLAr(Cl)] + [NiLAr(Cl)NHR] 2 pts

]
][
[
]
][
[
]
][
[
]
[
]
'
][
[
]
'
[
4
2
2

3
2
1
1

0 NiL ArCl

k
k
NaOBu
R
NH
ArCl
NiL
k
k
NiL
L
NiL
k
k

NiLL = − + + +

⎥
⎦
⎤
⎢
⎣
⎡

+
+
+
= − [ ]
]
][
[
]
[
1
]
'
[
]
[
]
'
[
4
2
2
3
2
1
1
0 ArCl
k
k
NaOBu
R

NH
ArCl
k
k
L
k
k
NiL

NiLL 3 pts

]
][
][
[
]
[
]
][
[
]
][
][
'
[
]
][
[
]
'

[
]
[
2
3
2
1
4
2
1
2
4
3
1
2
4
3
1
2
4
3
1

0 k kk L NH R NaOBu kk k NH R NaOBu kk k ArCl kk k ArCl NH R NaOBu

NaOBu
R
NH
k
k

k
NiLL
NiL
+
+
+
×
=
−
Equation 6
Substituting Equation 6 into the differential rate for [ArCl]:

]
][
[
]
[

2 ArCl NiL

k
dt

ArCl
d

− , results in the following predicted rate law expression for
the reaction mechanism:

d[ArNHR]/dt = - d[ArCl]/dt =

k2[ArCl] [NiL] = k1k2k3k4 [ArCl][NiLL’]0[NaOtBu][NH2R]

/{k-1k3k4[NH2R][NaOBu][L’] + k1k3k4[NaOBu][NH2R] + k1k2k4[ArCl] + k1k2k3

</div>
(35)<div class='page_container' data-page=35>

6e.Give the simplified form of the rate equation in 6d assuming that k1 is very small.
d[ArNHR]/dt = - d[ArCl]/dt = k2[ArCl] [NiL] = k1k2 [ArCl][NiLL’]0 / k-1[L’]

</div>
(36)<div class='page_container' data-page=36>

Problem 7. Synthesis of Artemisinin

(+)-Artemisinin, isolated from Artemisia annua L.
(Qinghao, Compositae) is a potent antimalarial
effective against resistant strains of Plasmodium. A
simple route for the synthesis of Artemisinin is
outlined below.

First, pyrolysis of (+)-2-Carene broke the cyclopropane ring forming, among other
products, (1R)-(+)-trans-isolimonene A (C10H16), which then was subjected to

regioselective hydroboration using dicyclohexylborane to give the required alcohol B

in 82% yield as a mixture of diastereoisomers. In the next step, B was converted to the
corresponding γ,δ-unsaturated acid C in 80% yield by Jones’ oxidation.

7a.Draw the structures (with stereochemistry) of the compounds A-C.

A B C

Me
H
HO

4 pts (2 pts if wrong stereochemistry) 4 pts 4 pts

Code: Question 7a 7b 7c 7d 7e 7f Total

Examiner Mark 12 8 8 12 12 12 64

Theoretical

Problem 7 
8.0 % of the

</div>
(37)<div class='page_container' data-page=37>

The acid C was subjected to iodolactonization using KI, I2 in aqueous. NaHCO3

solution to afford diastereomeric iodolactones D and E (which differ in
stereochemistry only at C3 ) in 70% yield.

7b. Draw the structures (with stereochemistry) of the compounds D and E.

The acid C was converted to diastereomeric iodolactones D and E (epimeric at the chiral
center C3). Look at the number-indicated in the structure F in the next step.

D E

4 pts 4pts

The iodolactone D was subjected to an intermolecular radical reaction with ketone

X using tris(trimethylsilyl)silane (TTMSS) and AIBN (azobisisobutyronitrile) in a
catalytic amount, refluxing in toluene to yield the corresponding alkylated lactone F in
72% yield as a mixture of diastereoisomers which differ only in stereochemistry at C7

</div>
(38)<div class='page_container' data-page=38>

7c. Draw the structures (with stereochemistry) of compound H and the reagent X.

Because alkylated lactone F is known, we can deduce the reagent X as methyl vinyl
ketone. H is the reduced product of D.

X H

2 pts 6 pts

The keto group of F reacted with ethanedithiol and BF3•Et2O in dichloromethane

</div>
(39)<div class='page_container' data-page=39>

7d.Draw the structures (with stereochemistry) of the compounds I and J.

The keto group of lactone F reacted with ethanedithiol and BF3·Et2O in dichloromethane

to afford thioketal lactones, I and the major isomer J.

I J

6 pts (3 pts if I and J are swapped) 6 pts (3 pts if I and J are swapped)

The isomer J was further subjected to alkaline hydrolysis followed by
esterification with diazomethane providing hydroxy methyl ester K in 50% yield. The

hydroxy methyl ester K was transformed into the keto ester L using PCC (Pyridium

ChloroChromate) as the oxidizing agent in dichloromethane (DCM).

</div>
(40)<div class='page_container' data-page=40>

7e.Draw the structures (with stereochemistry) of the compounds K and L.

Hydrolysis followed by esterification of J provided hydroxy ester K.

Oxidation of the hydroxy group in K by PCC resulted in the keto ester L in which two
protons adjacent to the carbonyl group are cis-oriented.

K L

6 pts 6 pts

The ketone L was subjected to a Wittig reaction with methoxymethyl
triphenylphosphonium chloride and KHMDS (Potassium HexaMethylDiSilazid - a
strong, non-nucleophilic base) to furnish the required methyl vinyl ether M in 45%
yield. Deprotection of thioketal using HgCl2, CaCO3 resulted in the key intermediate N

(80%). Finally, the compound N was transformed into the target molecule Artemisinin
by photo-oxidation followed by acid hydrolysis with 70% HClO4.

L M N

KHMDS

Ph3P(Cl)CH2OCH3 HgCl2, CaCO3 1. O2, hυ

</div>
(41)<div class='page_container' data-page=41>

7f.Draw the structures (with stereochemistry) of the compounds M and N.

The Wittig reaction of the ketone L resulted in the formation of methyl vinyl ether M.
Deprotection of the thioketal group forms the intermediate N.

M N

</div>
(42)<div class='page_container' data-page=42>

Problem 8. Star Anise

Illicium verum, commonly called Star anise, is a small native evergreen tree grown
in northeast Vietnam. Star anise fruit is used in traditional Vietnamese medicine. It is
also a major ingredient in the making the flavour of ‘phở’, a Vietnamese favourite
soup.

Acid A is isolated from the star anise fruit. The constitutional formula of A has been
deduced from the following sequence of reactions:

(I): this overall process results in alkene cleavage at the C=C bond, with each carbon of this
becoming doubly bonded to an oxygen atom.

(II): this oxidative cleavage process of 1,2-diols breaks C(OH)–C(OH) bond and produces
corresponding carbonyl compounds.

8a.Draw the structures for the compounds Y1 and Y2 and hence deduce the structure

of Y3 and A, B, C, D, given that in A there is only one ethylenic hydrogen atom.

Y1 Y2 Y3

CH

OH

1pt 1 pt 2 pts

Code: Question 8a 8b 8c 8d Total

Examiner Marks 15 2 12 10 39

Theoretical

Problem 8 
8.0% of the

</div>
(43)<div class='page_container' data-page=43>

A B 
COOH

HO
OH

O
O

COOH

HO
OH

O
HO

O
or

2 pts 2 pts

C D

3 pts 4 pts

Anethole, a main component of star anise oil, is
an inexpensive chemical precursor for the production of
many pharmaceutical drugs.

Treating anethole with sodium nitrite in acetic acid gives a crystalline solid E

(C10H10N2O3). The IR spectrum of E shows there is no non-aromatic C=C double

bond. The 1H NMR spectrum of E is given below.

2H 2H

</div>
(44)<div class='page_container' data-page=44>

8b. What differences in the structure between E and anethole can be obtained from
the 1H NMR data?

i) E contains a cis-C=C ethylenic bond while that of anethole is trans.
ii) E cannot contain a non-aromatic C=C bond.

iii) E is the adduct of anethole and N2O2.

iv) E is the adduct of anethole and N2O3.

v) E does not contain two trans ethylenic protons as anethole.

Pick one of the above statements

From 1H NMR data v [only] 2 pts

On heating at 150 oC for several hours, E is partially isomerized into F. Under the
same conditions, F gives the identical equilibrium mixture to that obtained from E. On
heating with phosphorus trichloride, both E and F lose one oxygen atom giving
compound G.Compounds E and F have the same functional groups.

The chemical shifts of methyl protons in E, F and G are given below.

E F G

CH3-O 3.8 ppm 3.8 ppm 3.8 ppm

CH3-C 2.3 ppm 2.6 ppm 2.6 ppm

8c. Suggest structures for E, F and G, assuming that they do NOT contain
three-membered rings.

E F G

</div>
(45)<div class='page_container' data-page=45>

A simplified structure for compound E is shown below; the R group does not
change throughout the rest of this question. Compound E is nitrated and subsequently
reduced with sodium dithionite to H. Treatment of H with sodium nitrite and
hydrochloric acid at 0–5 oC and subsequently reduced with stannous chloride to
provide I (R–C7H9N2O). One-pot reaction (three component reaction) of H,

benzaldehyde and thioglycolic acid (HSCH2CO2H) leads to the formation of J.

</div>
(46)<div class='page_container' data-page=46>

8d. Give the structuresfor H, I, J and K.

H I

2 pts 2 pts

J K

</div>
(47)<div class='page_container' data-page=47>

Problem 9. Heterocycle Preparation

Tetramethylthiurame disulfide (TMTD) is emerging as a useful reagent to prepare
many sulfur-nitrogen functional groups and heterocycles in organic chemistry. The
reactions of TMTD with primary amines, as well some corresponding
post-transformations of the resulting product(s) are presented in the following schemes:

(1)

(2)

(3)

(4)

(5)

Similar transformations of benzohydrazides (containing nucleophilic NH group)

Code: Question 9a 9b 9c 9d 9e 9f Total

Examiner Marks 8 4 6 4 2 9 33

Theoretical

Problem 9 
7.5 % of the

</div>
(48)<div class='page_container' data-page=48>

In the synthetic scheme below, the thiocarbamoylation reaction of an aroyl
hydrazine with TMTD produces compound C containing a heterocyclic moiety from
p-aminobenzoic acid.

During the formation of C from B, an intermediate B' was observed. This intermediate
tautomerizes to B''. C can be formed from B' or B''.

C

N
H

O
N
H

N=C=S
Me2N

B'

B''

9a.Give the structures of A, B, and C.

A B C

</div>
(49)<div class='page_container' data-page=49>

9b. Suggest a structure for the tautomer B’’ and give a curly-arrow mechanism for
the formation of C.

4 pts

Compound C was then converted to F by the following pathway:

[The group R remains exactly the same throughout the rest of the question.]

9c.Draw the structures of E, and F. (You do not need to draw the structure for the R group
from this point)

E F

2 pts 4 pts

E was only obtained when D was slowly added to the solution of excess N2H4

in dioxane. If N2H4 was added to the solution of D in dioxane instead, a major side

</div>
(50)<div class='page_container' data-page=50>

4 pts

Slightly heating D with ethanolamine (HOCH2CH2NH2) in dioxane for 2 hours

yielded G (R–C9H11N2OS).

9e.Draw the structural formula of G.

G

2 pts

9f. Heating G in the presence of p-toluenesulfonic acid as the catalyst could form a
number of different five-membered heterocyclic products.

i) Draw 2 structures that have different molecular formulae.

Eg H1

(3 pts)

H3

</div>
(51)<div class='page_container' data-page=51>

E.g H1

(3 pts)

H5

iii) Draw 2 structures that are stereoisomers. (3 pts)

E.g H5

(3 pts)

H7

H1 H2 H3

2 pts 2 pts 2 pts

H4 H5 H6

2 pts 2 pts 2 pts

H7 H8

</div>
(52)<div class='page_container' data-page=52>

Periodic Table of the Elements

6 Lanthanides 
58 
Ce 
140.1
59 
Pr 
140.9
60 
Nd 
144.2
61 
Pm

(144.9)

62 
Sm 
150.4
63 
Eu 
152.0
64 
Gd 
157.3
65 
Tb 
158.9
66 
Dy 
162.5
67 
Ho 
164.9
68 
Er 
167.3
69 
Tm 
168.9
70 
Yb 
173.0
71 
Lu

174.0
7 Actinides 
90 
Th 
91 
Pa 
92 
U 
93 
Np 
94 
Pu 
95 
Am 
96 
Cm 
97 
Bk 
98 
Cf 
99 
Es 
100 
Fm 
101 
Md 
102 
No 
103 
Lr

1 18

1

H

1.008 2 13 14 15 16 17

2 
He 
4.003
2
3 
Li 
6.941
4 
Be

9.012 Transition Elements

5 
B 
10.81
6 
C 
12.01
7

N 
14.01
8 
O 
16.00
9 
F 
19.00
10 
Ne 
20.18
3
11 
Na 
22.99
12 
Mg

24.31 3 4 5 6 7 8 9 10 11 12

13 
Al 
26.98
14 
Si 
28.09
15 
P 
30.98
16

S 
32.07
17 
Cl 
35.45
18 
Ar 
39.95
4
19 
K 
39.10
20 
Ca 
40.08
21 
Sc 
44.96
22 
Ti 
47.87
23 
V 
50.94
24 
Cr 
52.00
25 
Mn 
54.94

26 
Fe 
55.85
27 
Co 
58.93
28 
Ni 
58.69
29 
Cu 
63.55
30 
Zn 
65.41
31 
Ga 
69.72
32 
Ge 
72.61
33 
As 
74.92
34 
Se 
78.96
35 
Br 
79.90

36 
Kr 
83.80
5
37 
Rb 
85.47
38 
Sr 
87.62
39 
Y 
88.91
40 
Zr 
91.22
41 
Nb 
92.91
42 
Mo 
95.94
43 
Tc 
(97.9)
44 
Ru 
101.1
45 
Rh

102.9
46 
Pd 
106.4
47 
Ag 
107.9
48 
Cd 
112.4
49 
In 
114.8
50 
Sn 
118.7
51 
Sb 
121.8
52 
Te 
127.6
53 
I 
126.9
54 
Xe 
131.3
6
55

Cs 
132.9
56 
Ba 
137.3
57 
La 
138.9
72 
Hf 
178.5
73 
Ta 
180.9
74 
W 
183.8
75 
Re 
186.2
76 
Os 
190.2
77 
Ir 
192.2
78 
Pt 
195.1
79

Au 
197.0
80 
Hg 
200.6
81 
Tl 
204.4
82 
Pb 
207.2
83 
Bi 
209.0
84 
Po

(209.0)

85

At

(210.0)

86

Rn

(222.0)

87

Fr

(223.0)

88

Ra

(226.0)

89

Ac

(227.0)

</div>

Đề thi Olympic Hóa học IChO năm 2014

<b>46</b>

<b>th</b>

<b> International Chemistry Olympiad </b>

<b>July 25, 2014 </b>

<b>Hanoi, Vietnam </b>

<b>THEORETICAL EXAMINATION </b>

WITH ANSWER SHEETS GRADING

<b>GENERAL INTRODUCTION </b>

[

]

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+

+

=

8

<i>L</i>

<i>n</i>

<i>L</i>

<i>n</i>

<i>L</i>

<i>n</i>

<i>m</i>

<i>h</i>

<i>E</i>

<b>X</b>

<b>Y</b>

<b>Z</b>

(

)

[ ]

[ ]

[ ]

[ ]

[ ]

[ ]

<b>Problem 6. Chemical Kinetics </b>

<b>Problem 8. Star Anise </b>

CH

OH

<b>Periodic Table of the Elements</b>

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